Three vectors P- Q and R are shown in the figure- Let S be any point on the vector R- The distance between the points P and S is b-R- The general relation among vectors P- Q and S is

The correct option is D S = (1 b)P+bQ From triangular law of vector addition, we get #160; OP+PS = OS ∴ #160; P⃗+b|R⃗|R⃗|R⃗|=S⃗ P ...

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Byju's Answer

Standard XII

Physics

Vector Addition

Three vectors...

Question

Three vectors $\to P, \to Q$ and $\to R$ are shown in the figure. Let S be any point on the vector $\to R$ . The distance between the points P and S is $b | \to R |$ . The general relation among vectors $\to P, \to Q$ and $\to S$ is

\to S = (1 - b) \to P + b^{2} \to Q

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\to S = (1 - b^{2}) \to P + b \to Q

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\to S = (1 - b) \to P + b \to Q

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\to S = (b - 1) \to P + b \to Q

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Solution

The correct option is D $\to S = (1 - b) \to P + b \to Q$
From triangular law of vector addition, we get $O P + P S = O S$
$∴$ $\to P + b | \to R | \frac{\to R}{| \to R |} = \to S$
$\to P + b \to R = \to S$
But $\to R = \to Q - \to P$ (Given)
$\to P + b (\to Q - \to P) = \to S$
$⟹$ $\to S = (1 - b) \to P + b \to Q$

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